已知α为锐角,且tanα=1/2,求(sin2αcos2α-sinα)/(sin2αcos2α) 的值要详细过程那!
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![已知α为锐角,且tanα=1/2,求(sin2αcos2α-sinα)/(sin2αcos2α) 的值要详细过程那!](/uploads/image/z/576127-55-7.jpg?t=%E5%B7%B2%E7%9F%A5%CE%B1%E4%B8%BA%E9%94%90%E8%A7%92%2C%E4%B8%94tan%CE%B1%EF%BC%9D1%2F2%2C%E6%B1%82%28sin2%CE%B1cos2%CE%B1-sin%CE%B1%29%2F%28sin2%CE%B1cos2%CE%B1%29+%E7%9A%84%E5%80%BC%E8%A6%81%E8%AF%A6%E7%BB%86%E8%BF%87%E7%A8%8B%E9%82%A3%21)
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已知α为锐角,且tanα=1/2,求(sin2αcos2α-sinα)/(sin2αcos2α) 的值要详细过程那!
已知α为锐角,且tanα=1/2,求(sin2αcos2α-sinα)/(sin2αcos2α) 的值
要详细过程那!
已知α为锐角,且tanα=1/2,求(sin2αcos2α-sinα)/(sin2αcos2α) 的值要详细过程那!
tanα=1/2===>sinα=1/√5,cosα=2/√5
∴原式=1-sinα/[2sinαcosα(1-2sin²α)]
=1-(1/√5)/[2(1/√5)(2/√5)(1-2/5)]
=1-(1/√5)/[(4/5)*(3/5)]
=1-(1/√5)(25/12)
=1-5√5/12